Question

Consider an arithmetic sequence of algebraic from 3n+2.
(a) calculate the sum of first 10 term of this sequence ​

Answer 1

Answer :

S(10) = 185

Note :

★ A.P. (Arithmetic Progression) : A sequence in which the difference between the consecutive terms are equal is said to be in A.P.

★ If a1 , a2 , a3 , . . . , an are in AP , then

a2 - a1 = a3 - a2 = a4 - a3 = . . .

★ The common difference of an AP is given by ; d = a(n) - a(n-1) .

★ The nth term of an AP is given by ;

a(n) = a1 + (n - 1)d .

★ If a , b , c are in AP , then 2b = a + c .

★ The sum of nth terms of an AP is given by ;

S(n) = (n/2)×[ 2a + (n - 1)d ] .

S(n) = (n/2)×(a + l) , where l is last term .

★ The nth term of an AP can be also given by ; a(n) = S(n) - S(n-1) .

★ A linear polynomial in variable n always represents the nth term of an AP .

★ A quadratic polynomial in variable n always represents the sum of n terms of an AP .

★ If each terms of an AP is multiplied or divided by same quantity , then the resulting sequence is an AP .

★ If same quantity is added or subtracted in each term of an AP then the resulting sequence is an AP .

Solution :

• Given : nth term , T(n) = 3n + 2
• To find : S(10) = ?

Here ,

The nth term of the AP is ,

T(n) = 3n + 2

Thus ,

First term , a = T(1)

=> a = 3×1 + 2

=> a = 3 + 2

=> a = 5

Here ,

We need to find sum of first 10 terms .

Thus , n = 10

Also ,

Last term , l = T(10)

=> l = 3×10 + 2

=> l = 30 + 2

=> l = 32

Now ,

We know that , sum of first n terms is given by ;

S(n) = (n/2)×(a + l)

Thus ,

Sum of first 10 terms will be ,

=> S(10) = (10/2)×(5 + 32)

=> S(10) = 5×37

=> S(10) = 185

Hence , S(10) = 185

Answer 2

Given ,

• The general formula of an AP is 3n + 2

Thus ,

First term = 3(1) + 2 = 5

Second term = 3(2) + 2 = 8

Third term = 3(3) + 2 = 11

Fourth term = 3(4) + 2 = 14

$\sf \therefore \underline{The \: common \: difference \: (d) \: is \: 3}$

Now , the sum of first n terms of an AP is given by

$\boxed{ \sf{S_{n} = \frac{n}{2} \{2a + (n - 1)d \}}}$

Thus ,

$\sf \mapsto S_{10} = \frac{10}{2} \{2 \times 5 + (10 - 1) \times 3 \} \\ \\ \sf \mapsto S_{10} =5 \{10 + 27 \} \\ \\ \sf \mapsto S_{10} = 5 \times 37 \\ \\ \sf \mapsto S_{10} = 185$

$\sf \therefore \underline{The \: sum \: of \: first \: ten \: terms \: of \: given \: AP \: is \: 185}$